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Sav [38]
3 years ago
8

Sections of this interstate have speed limits of 65mph. A construction site is located on this interstate near Boise, ID. In the

construction site, the speed limit is reduced to 20 mph. A warning sign reading "WORK ZONE: 20 MPH SPEED ZONE" is required to be put before the start of the construction site so as to allow drivers to decelerate before entering the work zone. The sign is located 20 ft. to the side of the road, the deceleration rate of the vehicle is 11 ft./sec2 (recommended maximum deceleration rate by AASHTO) and the driver is assumed to have 20/40 vision. You may assume a perception time of 1 sec, an interpretation rate of 3 words/sec, and a decision time of 0.5 sec. Assuming the road is level answer the following questions.
a. What should be the minimum distance away from the warning sign from where the driver is able to read the warning? (Assuming 10 degree cone of vision and the vehicle is moving on the side of the road i.e. lateral distance between car and warning sign is 20 ft.)

b. Determine the total time taken by driver before he applies brakes. Also, determine the distance travelled during perception, interpretation, and reaction.

c. Assuming that the driver reacts only once he is out of 10 degree cone of vision, calculate the minimum distance away from the start of the construction site where a warning sign must be located to allow drivers to decelerate before entering the work zone.

d. Determine the minimum size of the text on the sign.
Engineering
1 answer:
e-lub [12.9K]3 years ago
7 0

Answer:

Explanation:

Given Preception time = 1.0 sec

Deceleration rate = 11 ft /sec2

Interpretation rate as = 3words /sec

The board consists of “WORK ZONE: 20 MPH SPEED ZONE”

Then Interpretation time = 2.0 sec

Decision time = 0.5 sec

Assume that the vehicle was traveling with 65mph

Perception Time

So, if you’re driving at 65 mph, your vehicle will travel 71 feet before you realize you need to start braking.

Reaction Distance

At 65 mph, that’s another 71 feet traveled.

Braking Distance

At 65 mph, it takes an additional 5.5 seconds or about 525 feet of actual brake application to stop your vehicle.

Stopping Distance

To determine the stopping distance, you calculate: Perception Distance (71 feet) + Reaction Distance (71 feet) + Braking Distance (525 feet) = Stopping Distance (667 feet)

Then to reduce the speed it takes 7.0 secs

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A pin fin of uniform cross-sectional area is fabricated of an aluminum alloy (k = 160W m-1 K-1 ). The fin diameter is D = 4 mm,
disa [49]

Answer: (a) 36.18mm

(b) 23.52

Explanation: see attachment

4 0
4 years ago
You are evaluating the lifetime of a turbine blade. The blade is 4 cm long and there is a gap of 0.16 cm between the tip of the
Tcecarenko [31]

Answer:

Explanation:

Given conditions

1)The stress on the blade is 100 MPa

2)The yield strength of the blade is 175 MPa

3)The Young’s modulus for the blade is 50 GPa

4)The strain contributed by the primary creep regime (not including the initial elastic strain) was 0.25 % or 0.0025 strain, and this strain was realized in the first 4 hours.

5)The temperature of the blade is 800°C.

6)The formula for the creep rate in the steady-state regime is dε /dt = 1 x 10-5 σ4 exp (-2 eV/kT)

where: dε /dt is in cm/cm-hr σ is in MPa T is in Kelvink = 8.62 x 10-5 eV/K

Young Modulus, E = Stress, \sigma /Strain, ∈

initial Strain, \epsilon_i = \frac{\sigma}{E}

\epsilon_i = \frac{100\times 10^{6} Pa}{50\times 10^{9} Pa}

\epsilon_i = 0.002

creep rate in the steady state

\frac{\delta \epsilon}{\delta t} = (1 \times {10}^{-5})\sigma^4 exp^(\frac{-2eV}{kT} )

\frac{\epsilon_{initial} - \epsilon _{primary}}{t_{initial}-t_{final}} = 1 \times 10^{-5}(100)^{4}exp(\frac{-2eV}{8.62\times10^{-5}(\frac{eV}{K} )(800+273)K} )

but Tinitial = 0

\epsilon_{initial} - \epsilon _{primary}} = 0.002 - 0.003 = -0.001

\frac{-0.001}{-t_{final}} = 1 \times 10^{-5}(100)^{4}\times 10^{(\frac{-2eV}{8.62\times10^{-5}(\frac{eV}{K} )1073K} )}

solving the above equation,

we get

Tfinal = 2459.82 hr

3 0
3 years ago
A rod that was originally 100-cm-long experiences a strain of 82%. What is the new length of the rod?
Ierofanga [76]

Answer: (b)

Explanation:

Given

Original length of the rod is L=100\ cm

Strain experienced is \epsilon=82\%=0.82

Strain is the ratio of the change in length to the original length

\Rightarrow \epsilon =\dfrac{\Delta L}{L}\\\\\Rightarrow 0.82=\dfrac{\Delta L}{100}\\\\\Rightarrow \Delta L=82\ cm

Therefore, new length is given by (Considering the load is tensile in nature)

\Rightarrow L'=\Delta L+L\\\Rightarrow L'=82+100=182\ cm

Thus, option (b) is correct.

8 0
3 years ago
A wood pole with a diameter of 10 in. has a moisture content of 5%. The fiber saturation point (FSP) for this wood is 30%. The w
Mekhanik [1.2K]

Answer:

a) Δd(change in wood diameter) = 5%

b) The wood would swell since the moisture content is increasing which will also led to increase in the wood's diameter

C) new diameter (D2) = 10.5 in

Explanation:

Wood pole diameter = 10 inches

moisture content = 5%

FSP = 30%

A) The percentage change in the wood's diameter

note : moisture fluctuations from 5% to 30% causes dimensional changes in the wood but above 30% up to 55% causes no change. hence this formula can be used to calculate percentage change in the wood's diameter

Δd/d = 1/5(30 - 5)

Δd/d = 5%  

Δd = 5%

B) would the wood swell or shrink

The wood would swell since the moisture content is increasing which will also led to increase in the wood's diameter

C) The new diameter of the wood

D2 = D + D( \frac{M1}{100} )

D = initial diameter= 10 in , M1 = initial moisture content = 5%

therefore D2 = 10 + 10( 5/100 )

new diameter (D2) = 10.5 in

5 0
3 years ago
1 import java.util.Scanner; 3 public class EqualityAndRelational { 4 public static void main (String args) args) { int userBonus
Anastaziya [24]

Missing Part of the Question

Complete the expression so that userPoints is assigned with 0 if userBonus is greater than 20 (second branch). Otherwise, userPoints is assigned with 10 (first branch

import java.util.Scanner;

public class EqualityAndRelational {

public static void main (String args) args) {

int userBonus; int userPoints;

userPoints=0;

Scanner scnr = new Scanner(System.in);

userBonus = scnr.nextInt();

// Program will be tested with values : 15, 20, 25, 30, 35. 12 13 14 15 16 17 18 19

( Your solution goes here)

{

userPoints= 10 ;

}

else {

userPoints = 0;

}

}

}

Answer;

Replace

( Your solution goes here)

With

if(userBonus>20).

The full program becomes

import java.util.Scanner;

public class EqualityAndRelational {

public static void main (String args) args) {

int userBonus; int userPoints;

userPoints=0;

Scanner scnr = new Scanner(System.in);

userBonus = scnr.nextInt();

// Program will be tested with values : 15, 20, 25, 30, 35. 12 13 14 15 16 17 18 19

if(userBonus>20)

{

userPoints= 10 ;

}

else {

userPoints = 0;

}

}

}

7 0
4 years ago
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